Educational Codeforces Round 53 (Rated for Div. 2) C Vasya and Robot 二分

题目：题目链接

思路：对于x方向距离与y方向距离之和大于n的情况是肯定不能到达的，另外，如果n比abs(x) + abs(y)大，那么我们总可以用UD或者LR来抵消多余的大小，所以只要abs(x) + abs(y) <= n && (n - abs(x) + abs(y)) % 2 == 0，就一定可以到达终点，判断完之后二分答案长度就可以了

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <list>
#include <cstdlib>
#include <cmath>

#define FRER() freopen("in.txt", "r", stdin)
#define FREO() freopen("out.txt", "w", stdout)
#define INF 0x3f3f3f3f

using namespace std;

const int maxn = 200000 + 5;

int n, cx[maxn], cy[maxn], x, y;
char str[maxn];

bool judge(int t) {
    for(int i = t; i <= n; ++i) {
        int lx = cx[n] - cx[i] + cx[i - t];
        int ly = cy[n] - cy[i] + cy[i - t];
        if (abs(x - lx) + abs(y - ly) <= t)
            return true;
    }
    return false;
}

int main()
{
    cin >> n >> str >> x >> y;
    if(abs(x) + abs(y) > n || (abs(x) + abs(y)) % 2 != n % 2) {
        cout << -1 << endl;
    }
    else {
        for(int i = 0; i < n; ++i) {
            if(str[i] == 'U') cx[i + 1] = cx[i], cy[i + 1] = cy[i] + 1;
            else if(str[i] == 'D') cx[i + 1] = cx[i], cy[i + 1] = cy[i] - 1;
            else if(str[i] == 'L') cx[i + 1] = cx[i] - 1, cy[i + 1] = cy[i];
            else if(str[i] == 'R') cx[i + 1] = cx[i] + 1, cy[i + 1] = cy[i];
        }
        int l = 0, r = n;
        while(l < r) {
            int m = (l + r) >> 1;
            if(judge(m))
                r = m;
            else
                l = m + 1;
        }
        cout << r << endl;
    }
    return 0;
}